$解:(1)在Rt△ABC中,由勾股定理得:$
$AB²=BC²+AC²$
$即AB=\sqrt{AC²+BD²}=\sqrt{5²+12²}=13$
$(2)S_{△ABC}=\frac {1}{2}AC·BC=\frac {1}{2}AB·CO$
$即CO=\frac {AC·BC}{AB}=\frac {60}{13}$
$在Rt△ACO中,由勾股定理得:$
$AC²=AO²+CO²$
$即AO=\sqrt{AC²-CO²}=\sqrt{5²-(\frac {60}{13})²}=\frac {25}{13}$